Alice and Bob
Time Limit: 1000ms Memory limit: 65536K 有疑問?點這裏^_^
題目描述
Alice and Bob like playing games very much.Today, they introduce a new game.
There is a polynomial like this: (a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1). Then Alice ask Bob Q questions. In the expansion of the Polynomial, Given an integer P, please tell the coefficient of the x^P.
Can you help Bob answer these questions?
輸入
For each case, the first line contains a number n, then n numbers a0, a1, .... an-1 followed in the next line. In the third line is a number Q, and then following Q numbers P.
1 <= T <= 20
1 <= n <= 50
0 <= ai <= 100
Q <= 1000
0 <= P <= 1234567898765432
輸出
示例輸入
122 1234
示例輸出
20
提示
來源
示例程序
#include<iostream>
#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
int a[70];
int b[70];
int main()
{
int i,j,k,q,n;
long long p;
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
memset(a,0,sizeof(a));//很重要
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
scanf("%d",&q);
int na=0;
int nb=0;
while(q--)
{
scanf("%lld",&p);
if(p==0)
{
printf("1\n");
continue;
}
na=0;
nb=1;
while(p)
{
b[na]=p%2;
na++;
p=p/2;
}
/*for(i=0;i<na;i++)
{
printf("bb %d\n",b[i]);
}*/
for(i=0;i<na;i++)
{
if(b[i])
{
nb=nb*a[i]%2012;
}
}
printf("%d\n",nb);
}
}
return 0;
}